LeetCode之1 bit and 2 bit Characters（Kotlin）

问题： We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11).

Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.

Note: 1 <= len(bits) <= 1000. bits[i] is always 0 or 1.

---

方法： 遍历数组，若是当前bit为0则指针移动一位，由于0必为first character；若是当前bit为1则指针移动两位，由于1X必为second character。当遍历到数组尾端时，若是指针正好等于数组最后一个元素的下标，则返回true；不然，数组尾端必为10，返回false。

具体实现：

class IsOneBitCharacter {
    fun isOneBitCharacter(bits: IntArray): Boolean {
        var i = 0
        while (i <= bits.lastIndex - 1) {
            if (bits[i] == 0) {
                i++
            } else if (bits[i] == 1) {
                i += 2
            }
        }
        if (i == bits.lastIndex) {
            return true
        }
        return false
    }
}

fun main(args: Array<String>) {
    val array = intArrayOf(0, 0, 0, 0)
    val isOneBitCharacter = IsOneBitCharacter()
    val result = isOneBitCharacter.isOneBitCharacter(array)
    println("result: $result")
}
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具体代码实现能够参考Github